jmc

Let $r,$ $s,$ and $t$ be the three roots, and suppose that $rs=30.$ By Vieta's formulas, we also know that $$\begin{array}{rl}r+s+t& =10,\\ rs+st+rt& =55.\end{array}$$Since $rs=30,$ the second equation simplifies to $st+rt=25,$ or $t(r+s)=25.$ Therefore, the numbers $r+s$ and $t$ have a sum of $10$ and a product of $25,$ so they must be the roots of the quadratic $$y^2-10y+25=0.$$This quadratic factors as $(y-5{)}^2=0,$ so we have $r+s=t=5.$ Then by Vieta again, $$c=rst=30\cdot 5=\boxed{150}.$$